The value of ΔH° for the reaction below is -126 kJ. ________ kj are released when 2.00 mol of NaOH is formed in the reaction?
2Na₂O₂ (s) + 2H₂O (l) → 4NaOH (s) + O₂ (g)

Question

Quizplus · Accepted Answer

The given ΔH° value (-126 kJ) is for the formation of 4 moles of NaOH. Therefore, to find the energy released for the formation of 2 mol of NaOH, we need to use stoichiometry and proportionality. 
ΔH° = -126 kJ is released for the formation of 4 mol of NaOH.
Therefore, ΔH°/4 = -31.5 kJ is released for the formation of 1 mol of NaOH.
So, for the formation of 2 mol of NaOH, the energy released would be:
2 mol × (-31.5 kJ/mol) = -63 kJ
Hence, the answer is B. 63.

The Value of ΔH° for the Reaction Below Is -126